Friday, March 2, 2018

Question: Could you explain the concept of a binary star system and how we get its orbit parameters correctly oriented to our view? (Since the system is not likely to be in the plane of our viewing axis initially) Also, how might one find the mass of each, say for a specified binary system?

Answer:

Below I sketch a basic binary star system for reference:

The
diagram above shows the simplest type of system, with the plane of the binary components A (of
mass m(A)) and B (of mass m(B)) perpendicular to the viewing plane. This visual binary system, it's important to
note, indicates the apparent relative orbit of the 2 stars - since ordinarily (as you pointed out) - the plane of the real orbit will not lie in the plane of the sky, i.e. perpendicular to the line of sight as portrayed. Usually, what we do is
observe the motions of the fainter member about the brighter over a period of
time sufficient to determine the orbit period, and then obtain the apparent
relative orbit.

Geometrically, it is straightforward to show that an ellipse
in one plane when projected onto another plane (say oblique to it) will yield
another ellipse but of different eccentricity, e = c/a. Most importantly, the
foci of the original ellipse do not project onto the foci of the projected one.
This means the primary (brightest star) though it is located at one focus of
the true relative orbit, is not at the focus of the apparent relative orbit.

It
is this circumstance which makes it possible to determine the inclination of
the true orbit to the plane of the sky. Basically, the problem reduces to
finding the angle at which the true relative orbit must be projected in order
to account for the amount of displacement of the primary from the focus of the
apparent relative orbit.

If the semi-major axis of the true
relative orbit (e.g. the one it would have if displayed face-on) has an angular
distance of a" (seconds of arc) and if the system is at a distance d
parsecs, then the semi-major axis in astronomical units is:

a = (a" x d)

Then the sum of the masses of the two stars is given by Kepler's 3rd law:

m(A) + m(B) = (d a")3/ P2

where P is their period.

Of
course, obtaining the total mass is only the first step. One then wishes to
obtain the individual masses for each star. This is done by analyzing the
motions of each member with respect to the center of mass of the system which
ordinarily will be apart from either member. For example, if m(A) = m(B) then
the center of mass (cm) will be exactly midway between them.

Consider
the visual binary system for Sirius composed of Sirius A and B. The semi-major axis of the true
relative orbit is observed to be 7½" and the distance from the Sun to Sirius is 2.67 pc.
If the period of the Sirius binary system is 50 years and the component B is
found to be twice as far from the center of mass as component A, then we can compute the mass of each component.We
first obtain the total mass from Kepler's 3rd law:

Spectroscopic
binaries are also of much interest and derive their name because spectroscopic
analysis is needed to obtain the radial velocity curve for the system and hence
the relative orbital velocity, V, for the pair. The distance around the
relative orbit, i.e. its circumference, is just the relative orbital velocity V
(deduced from the radial velocity profile) multiplied by the period P. Then the
distance between the stars a, is just:

a = (V x P)/ 2π

If, for instance, the relative velocity
is a lower limit, then the separation we obtain is a lower limit for the
system. If this is then applied to Kepler's 3rd law one can obtain a lower
limit to the sum of the masses of the components:

m1 + m2 = a3/P2

To fix ideas, say a spectroscopic
binary system is found to have a relative velocity of 100 km/sec and a period
of 17.5 days. Then let component
(1) be 3 times the mass of component (2) and find: a) a
lower limit for the separation of the components, a, and thence a lower limit
to the sum of the masses, and b) the
lower limits on the masses of the components.

First convert 100
km/sec to AU/yr. (Astronomical units per year)

Over one year: t = 3.156 x 107 s

total distance covered:

d = v x t = 100 km/s
(3.156 x 10 7 s) = 3.156 x 109 km

But 1 AU = 1.495 x 108 km

Then, the AU in this total distance:

d/AU = (3.156 x 109 km)/ (1.495 x 108 km) = 21.1 AU

The period in yrs. for 17.5 days:

P = 17.5/ (365.25) = 0.048 yr.

Then:

a = (21.1 AU x 0.048 yr/AU)/ 2π = 0.161 AU

The lower limit to the masses is therefore:

m1 + m2 = (0.161)3/ (0.048)2 = 1.8 solar masses

Since star m1 has
3x the mass of star m2, then: m1
= 3m2

And: m2 + 3m2 =
1.8 or 4m2 = 1.8

so:

m2 = 1.8/4 =
0.45 solar masses, and

m1 = 3(0.45) = 1.35 solar masses

Basically, in working any binary star problem, say to get the relative masses, one needs to take care the units used are consistent, and also that Kepler's 3rd law is applied in the correct format.

About Me

Specialized in space physics and solar physics, developed first astronomy curriculum for Caribbean secondary schools, has written thirteen books - the most recent:Fundamentals of Solar Physics. Also: Modern Physics: Notes, Problems and Solutions;:'Beyond Atheism, Beyond God', Astronomy & Astrophysics: Notes, Problems and Solutions', 'Physics Notes for Advanced Level&#39, Mathematical Excursions in Brane Space, Selected Analyses in Solar Flare Plasma Dynamics; and 'A History of Caribbean Secondary School Astronomy'. It details the background to my development and implementation of the first ever astronomy curriculum for secondary schools in the Caribbean.